Issue in understanding function pointer in C - c

Question

I am having problem in understanding function pointer for passing any arguments. Please find the sample program for your reference. Here I want to print the sum of two numbers. print() accepts a function pointer as its argument. But I am not able to pass the argument from main() to the function.
#include<stdio.h>
int sum(int, int);
void print(int (*p)(int,int));
int main()
{
int a=2;
int b=5;
print(sum(a,b));
}
void print(int (*p)(int a,int b))
{
printf("%d",p(a,b)); //a and b are not defined as per compiler
}
int sum(int a, int b)
{
return a+b;
}

Answer 1

When you make this declaration
void print(int (*p)(int a,int b))
you declare a function print that takes a function pointer p that takes two ints and return an int. The names a and b do not represent anything in particular - the compiler ignores them. In fact, you could omit them:
void print(int (*p)(int,int))
If you would like to pass two numbers to be added, pass them as separate parameters:
void print(int (*p)(int,int), int a, int b)
Now your print will compile. The call will need to look like this:
print(sum, a, b);
This is because when you write print(sum(a, b)) you are instructing the compiler to call sum with parameters a and b, obtain its result, and pass to print. However, when you pass sum by itself, with no parentheses, it represents the function pointer.

Answer 2

The program will look like
#include<stdio.h>
int sum( int, int );
void print( int ( * )( int, int ), int, int );
// or more simpler
void print( int ( int, int ), int, int );
int main()
{
int a = 2;
int b = 5;
print( sum, a, b);
}
void print( int ( *p )( int,int ), int a, int b )
{
printf( "%d", p( a, b ) ); //a and b are not defined as per compiler
}
int sum( int a, int b )
{
return a + b;
}
Take into account that these declarations are equivalent and declare the same function
void print( int ( * )( int, int ), int, int );
// or more simpler
void print( int ( int, int ), int, int );
As for your original code then function print only has one parameter: a function pointer. So you may pass to function print only one argument: some function. You could define function print the following way
void print( int ( *p )( int,int ) )
{
int a = 2, b = 5;
printf( "%d", p( a, b ) ); //a and b are not defined as per compiler
}
Or you can define function print with three parameters that the user of the function could himself specify arguments to the called function within the body of print.
void print( int ( *p )( int,int ), int a, int b )
{
printf( "%d", p( a, b ) ); //a and b are not defined as per compiler
}

Answer 3

Let's start with a basic function which we will be pointing to:
int addInt(int n, int m) {
return n+m;
}
First thing, lets define a pointer to a function which receives 2 ints and returns and int:
int (*functionPtr)(int,int);
Now we can safely point to our function:
functionPtr = &addInt;
Now that we have a pointer to the function, lets use it:
int sum = (*functionPtr)(2, 3); // sum == 5
Passing the pointer to another function is basically the same:
int add2to3(int (*functionPtr)(int, int)) {
return (*functionPtr)(2, 3);
}
We can use function pointers in return values as well (try to keep up, it gets messy):
// this is a function called functionFactory which receives parameter n
// and returns a pointer to another function which receives two ints
// and it returns another int
int (*functionFactory(int n))(int, int) {
printf("Got parameter %d", n);
int (*functionPtr)(int,int) = &addInt;
return functionPtr;
}
But it's much nicer to use a typedef:
typedef int (*myFuncDef)(int, int);
// note that the typedef name is indeed myFuncDef
myFuncDef functionFactory(int n) {
printf("Got parameter %d", n);
myFuncDef functionPtr = &addInt;
return functionPtr;
}
Disclaimer: this code was not necessarily compiled, if you see any errors, feel free to edit.